Lucas-Lehmer test

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Revision as of 23:13, 18 February 2008 by MikeMol (talk | contribs) (C++ version)
Task
Lucas-Lehmer test
You are encouraged to solve this task according to the task description, using any language you may know.

Lucas-Lehmer Test: for p a prime, the Mersenne number 2**p-1 is prime if and only if 2**p-1 divides S(p-1) where S(n+1)=S(n)**2-2, and S(1)=4.

The following programs calculate all Mersenne primes up to the implementation's maximium precision, or the 45th Mersenne prime. (Which ever comes first).

Ada

with Ada.Text_Io; use Ada.Text_Io;
with Ada.Integer_Text_Io; use Ada.Integer_Text_Io;

procedure Lucas_Lehmer_Test is
   type Ull is mod 2**64;
   function Mersenne(Item : Integer) return Boolean is
      S : Ull := 4;
      MP : Ull := 2**Item - 1;
   begin
      if Item = 2 then
         return True;
      else
         for I in 3..Item loop
            S := (S * S - 2) mod MP;
         end loop;
         return S = 0;
      end if;
   end Mersenne;
   Upper_Bound : constant Integer := 64;
begin
   Put_Line(" Mersenne primes:");
   for P in 2..Upper_Bound loop
      if Mersenne(P) then
         Put(" M");
         Put(Item => P, Width => 1);
      end if;
   end loop;
end Lucas_Lehmer_Test;

Output: 

Mersenne primes:
M2 M3 M5 M7 M13 M17 M19 M31

ALGOL 68

Interpretor: algol68g-mk11 

main:(
  PRAGMAT stack=1M precision=20000 PRAGMAT

  PROC is prime = ( INT p )BOOL:
    IF p = 2 THEN TRUE
    ELIF p <= 1 OR p MOD 2 = 0 THEN FALSE
    ELSE
      BOOL prime := TRUE;
      FOR i FROM 3 BY 2 TO ENTIER sqrt(p)
        WHILE prime := p MOD i /= 0 DO SKIP OD;
      prime
    FI;

  PROC is mersenne prime = ( INT p )BOOL:
    IF p = 2 THEN TRUE
    ELSE
      LONG LONG INT m p :=  LONG LONG 2 ** p - 1, s := 4;
      FROM 3 TO p DO
        s := (s ** 2 - 2) MOD m p
      OD;
      s = 0
    FI;

  INT upb prime = ( long long bits width - 1 ) OVER 2; # no unsigned #
  INT upb count = 45; # find 45 mprimes if INT has enough bits #

  printf(($" Finding Mersenne primes in M[2.."g(0)"]: "l$,upb prime));

  INT count:=0;
  FOR p FROM 2 TO upb prime WHILE
    IF is prime(p) THEN
      IF is mersenne prime(p) THEN
        printf (($" M"g(0)$,p));
        count +:= 1
      FI
    FI;
    count <= upb count
  DO SKIP OD
)

Output: 

Finding Mersenne primes in M[2..33252]: 
M2 M3 M5 M7 M13 M17 M19 M31 M61 M89 M107 M127 M521 M607 M1279 M2203 M2281 M3217 M4253 M4423 M9689 M9941 M11213 M19937 M21701 M23209

See also: http://www.xs4all.nl/~jmvdveer/mersenne.a68.html

C

#include <math.h>
#include <stdio.h>
#include <limits.h>
#pragma precision=log10l(ULLONG_MAX)/2

typedef enum { FALSE=0, TRUE=1 } BOOL;

BOOL is_prime( int p ){
  if( p == 2 ) return TRUE;
  else if( p <= 1 || p % 2 == 0 ) return FALSE;
  else {
    BOOL prime = TRUE;
    const int to = sqrt(p);
    int i;
    for(i = 3; i <= to; i+=2)  
      if (!(prime = p % i))break;
    return prime;
  }
}

BOOL is_mersenne_prime( int p ){
  if( p == 2 ) return TRUE;
  else {
    const long long unsigned m_p = ( 1LLU << p ) - 1;
    long long unsigned s = 4;
    int i;
    for (i = 3; i <= p; i++){
      s = (s * s - 2) % m_p;
    }
    return s == 0;
  }
}

int main(int argc, char **argv){

  const int upb = log2l(ULLONG_MAX)/2; 
  int p;
  printf(" Mersenne primes:\n");
  for( p = 2; p <= upb; p += 1 ){
    if( is_prime(p) && is_mersenne_prime(p) ){
      printf (" M%u",p);
    }
  }
  printf("\n");
}

Compiler version: gcc version 4.1.2 20070925 (Red Hat 4.1.2-27)
Compiler options: gcc -std=c99 -lm Lucas-Lehmer_test.c -o Lucas-Lehmer_test
Output: 

Mersenne primes:
M2 M3 M5 M7 M13 M17 M19 M31

C++

Library: GMP
#include <iostream>
#include <gmpxx.h>

mpz_class pow2(mpz_class exp);
bool is_mersenne_prime(mpz_class p);

int main()
{
	mpz_class maxcount(45);
	mpz_class found(0);
	mpz_class check(0);
	for( mpz_nextprime(check.get_mpz_t(), check.get_mpz_t());
	     found < maxcount;
	     mpz_nextprime(check.get_mpz_t(), check.get_mpz_t()))
	{
		//std::cout << "P" << check << " " << std::flush;
		if( is_mersenne_prime(check) )
		{
			++found;
			std::cout << "M" << check << " " << std::flush;
		}
	}
}

bool is_mersenne_prime(mpz_class p)
{
	if( 2 == p )
		return true;
	else
	{
		mpz_class div = pow2(p) - mpz_class(1);
		mpz_class s(4);
		
		for( mpz_class i(3);
			 i <= p;
			 ++i )
		{
			s =  (s * s - mpz_class(2)) % div ;
		}

		return ( s == mpz_class(0) );
	}
}

mpz_class pow2(mpz_class exp)
{
	// Unfortunately, GMP doesn't have a left-shift method.
	// It also doesn't have a pow() equivalent that takes arbitrary-precision exponents.
	// So we have to do it the hard (and presumably slow) way.
	mpz_class ret(2);
	
	for(mpz_class i(1); i < exp; ++i)
		ret *= mpz_class(2);
	
	//std::cout << "pow2( " << exp << " ) = " << ret << std::endl;
	return ret; 
}

Output: (Incomplete; It takes a long time.) 

M2 M3 M5 M7 M13 M17 M19 M31 M61 M89 M107 M127 M521 M607 M1279 M2203 M2281 M3217 M4253 M4423 M9689 M9941

Java

We use arbitrary-precision integers in order to be able to test any arbitrary prime. 

import java.math.BigInteger;
public class Mersenne
{

    public static boolean isPrime(int p) {
        if (p == 2)
            return true;
        else if (p <= 1 || p % 2 == 0)
            return false;
        else {
            int to = (int)Math.sqrt(p);
            for (int i = 3; i <= to; i += 2)
                if (p % i == 0)
                    return false;
            return true;
        }
    }

    public static boolean isMersennePrime(int p) {
        if (p == 2)
            return true;
        else {
            BigInteger m_p = BigInteger.ONE.shiftLeft(p).subtract(BigInteger.ONE);
            BigInteger s = BigInteger.valueOf(4);
            for (int i = 3; i <= p; i++)
                s = s.multiply(s).subtract(BigInteger.valueOf(2)).mod(m_p);
            return s.equals(BigInteger.ZERO);
        }
    }

    // an arbitrary upper bound can be given as an argument
    public static void main(String[] args) {
        int upb;
        if (args.length == 0)
            upb = 500;
        else
            upb = Integer.parseInt(args[0]);

        System.out.println(" Finding Mersenne primes in M[2.." + upb + "]: ");
        for (int p = 2; p <= upb; p++)
            if (isPrime(p) && isMersennePrime(p))
                System.out.print(" M" + p);
        System.out.println();
    }
}

Output: 

Finding Mersenne primes in M[2..500]:
M2 M3 M5 M7 M13 M17 M19 M31 M61 M89 M107 M127

Python

from sys import stdout
from math import sqrt, log
fi = od = None

def is_prime ( p ):
  if p == 2: return True
  elif p <= 1 or p % 2 == 0: return False
  else:
    prime = True;
    for i in range(3, int(sqrt(p))+1, 2 ): # do
      if p % i == 0: return False
    else:
      return True
    fi
  fi;

def is_mersenne_prime ( p ):
  if p == 2: return True
  else:
    m_p =  ( 1 << p ) - 1; s = 4;
    for i in range(3, p+1): # do
      s = (s ** 2 - 2) % m_p
    od;
    return s == 0
  fi;

precision = 20000; # maximum requested number of decimal places of 2 ** MP-1 #
long_bits_width = precision / log(2) * log(10);
upb_prime = int( long_bits_width - 1 ) / 2; # no unsigned #
upb_count = 45; # find 45 mprimes if int was given enough bits #

print ((" Finding Mersenne primes in M[2..%d]:"%upb_prime));

count=0;
for p in range(2, upb_prime+1): # do
  if is_prime(p) and is_mersenne_prime(p):
    print("M%d"%p),
    stdout.flush();
    count += 1
  fi;
  if count >= upb_count: break
od
print

Output: 

Finding Mersenne primes in M[2..33218]: 
M2 M3 M5 M7 M13 M17 M19 M31 M61 M89 M107 M127 M521 M607 M1279 M2203 M2281 M3217 M4253 M4423 M9689 M9941 M11213 M19937 M21701 M23209
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